# The graph of a quadratic function has a y intercept at 0,5 and a minimum at 3,-4 ?

Apr 25, 2018

$f \left(x\right) = {x}^{2} - 6 x + 5$

#### Explanation:

$f \left(x\right) = a {x}^{2} + b x + c$

$5 = f \left(0\right) = a \left({0}^{2}\right) + b \left(0\right) + c$

$c = 5$

The minimum $y$ is at $x = - \frac{b}{2 a} .$

$- \frac{b}{2 a} = 3$

$b = - 6 a$

$\left(3 , - 4\right)$ is on the curve:

$- 4 = f \left(3\right) = a {\left(3\right)}^{2} + \left(- 6 a\right) \left(3\right) + 5$

$- 9 = - 9 a$

$a = 1$

$b = - 6 a = - 6$

$f \left(x\right) = {x}^{2} - 6 x + 5$

Check: f(0)=5 quad sqrt

$f \left(x\right) = \left({x}^{2} - 6 x + 9\right) - 9 + 5 = {\left(x - 3\right)}^{2} - 4$ so $\left(3 , - 4\right)$ is the vertex.quad sqrt

Apr 25, 2018

$y = {\left(x - 3\right)}^{2} - 4$

#### Explanation:

Assuming that the equation of such quadratic graph is requested:

$y = a {\left(x - h\right)}^{2} + k$ => Equation of parabola in vertex form where:

$\left(h , k\right)$ is the vertex, for $a > 0$ the parabola opens up which

makes the vertex the minimum, so in this case $\left(3 , - 4\right)$ is the

vertex then:

$y = a {\left(x - 3\right)}^{2} - 4$ => the $y$ intercept is at: $\left(0 , 5\right)$:

$5 = a {\left(0 - 3\right)}^{2} - 4$ => solving for $a$:

$5 = 9 a - 4$

$9 = 9 a$

$a = 1$

Thus the equation of the graph is:

$y = {\left(x - 3\right)}^{2} - 4$